Aptitude - Probability - Discussion

As we know one dice has 6 sides(4 edges + 1frontside + 1backside = 6)
Possibilities of getting sum 9 are (6,3)(5,4)(4,5)(3,6) i.e 4 possibilities.
Because;
6+3 = 9
5+4 = 9
4+5 = 9
3+6 = 9.

Remember! The numbers you add should not exceed the 6 digit.

Solution for your question is @Dhaval.
4/6 * 6 * 6 = 4/216 or 1/54.
Hope,This Helps!

Here how could you consider (5, 4) and (4, 5) two different events? Please explain.

Why not taking the case {3,3}? Please explain me.

@Pooja Gm.

Dice does not have 8 surfaces it has only 6.

One dice have six faces so here we have two dices, so the formula is n2 (n Square). So 6*6 =36.

Is throwing a dice two times and throwing two dice together, same?

Please, anyone, clear me.

At first you draw sample space then you find pair which sum is 9, (1,1), (1,2),....... (5,4), (6,3),(3,6), (4,5).

Total pair 36 and sum of 9 pair is 4 them p(sum 9)= 4 ÷ 36 = 1/9.

Because in a throw of two dices there is only 6x6, is the last the number won't Or should not exceed 6.

Why we should not use the permutations here?like sum 9 can appear in either (6, 3) or (3, 6) , (5, 4) or (4, 5). Why we only choose the one possibility?

Why we take this case only?
(3+6)
(4+5)
(5+4)
(6+3).

This also possible;
(7+2),
(1+8),
(2+7),
(8+1),

Please explain the reason.